A2 June 2023 Paper 2 Q9
9. A patient is treated by administering an antibiotic intravenously at a constant rate for some time.
Initially there is none of the antibiotic in the patient.
At time \(t\) minutes after treatment began
- the concentration of the antibiotic in the blood of the patient is \(x\) mg/ml
- the concentration of the antibiotic in the tissue of the patient is \(y\) mg/ml
The concentration of antibiotic in the patient is modelled by the equations
\[\frac{\mathrm{d}x}{\mathrm{d}t} = 0.025y - 0.045x + 2\]\[\frac{\mathrm{d}y}{\mathrm{d}t} = 0.032x - 0.025y\]To be effective for the patient the concentration of antibiotic in the tissue must eventually reach a level between 185 mg/ml and 200 mg/ml.
| Scheme | Marks | AO |
|---|---|---|
\(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}t^2} = 0.032\frac{\mathrm{d}x}{\mathrm{d}t} - 0.025\frac{\mathrm{d}y}{\mathrm{d}t}\) \(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}t^2} = \frac{4}{125}\frac{\mathrm{d}x}{\mathrm{d}t} - \frac{1}{40}\frac{\mathrm{d}y}{\mathrm{d}t}\) \(\displaystyle \frac{\mathrm{d}x}{\mathrm{d}t} = \frac{1}{0.032}\left(0.025\frac{\mathrm{d}y}{\mathrm{d}t} + \frac{\mathrm{d}^2y}{\mathrm{d}t^2}\right)\) \(\displaystyle \frac{\mathrm{d}x}{\mathrm{d}t} = \frac{25}{32}\frac{\mathrm{d}y}{\mathrm{d}t} + \frac{125}{4}\frac{\mathrm{d}^2y}{\mathrm{d}t^2}\) | B1 | 1.1b |
\(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}t^2} = 0.032(0.025y - 0.045x + 2) - 0.025\frac{\mathrm{d}y}{\mathrm{d}t}\) \(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}t^2} = 0.0008y - 0.00144x + 0.064 - 0.025\frac{\mathrm{d}y}{\mathrm{d}t}\) \(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}t^2} = \frac{1}{1250}y - \frac{9}{6250}x + \frac{8}{125} - \frac{1}{40}\frac{\mathrm{d}y}{\mathrm{d}t}\) Then substitutes for \(x\) \(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}t^2} = 0.0008y - \frac{0.00144}{0.032}\left(\frac{\mathrm{d}y}{\mathrm{d}t} + 0.025y\right) + 0.064 - 0.025\frac{\mathrm{d}y}{\mathrm{d}t}\) or \(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}t^2} = \frac{1}{1250}y - \frac{9}{6250}\left(\frac{125}{4}\frac{\mathrm{d}y}{\mathrm{d}t} + \frac{25}{32}y\right) + \frac{8}{125} - \frac{1}{40}\frac{\mathrm{d}y}{\mathrm{d}t}\) or \(\displaystyle \frac{1}{0.032}\left(\frac{\mathrm{d}^2y}{\mathrm{d}t^2} + 0.025\frac{\mathrm{d}y}{\mathrm{d}t}\right) = 0.025y - \frac{0.045}{0.032}\left(\frac{\mathrm{d}y}{\mathrm{d}t} + 0.025y\right) + 2\) | M1 | 1.1b |
\(\displaystyle\left\{\frac{\mathrm{d}^2y}{\mathrm{d}t^2} = -0.000325y - 0.07\frac{\mathrm{d}y}{\mathrm{d}t} + 0.064\right\}\) \(\displaystyle\left\{\frac{\mathrm{d}^2y}{\mathrm{d}t^2} = -\frac{13}{40000}y - \frac{7}{100}\frac{\mathrm{d}y}{\mathrm{d}t} + \frac{8}{125}\right\}\) \(\displaystyle 40\,000\frac{\mathrm{d}^2y}{\mathrm{d}t^2} + 2800\frac{\mathrm{d}y}{\mathrm{d}t} + 13y = 2560\,*\) | A1* | 2.1 |
| (3) |
Notes
Alternative
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}t^2} = 0.032\frac{\mathrm{d}x}{\mathrm{d}t} - 0.025\frac{\mathrm{d}y}{\mathrm{d}t}\) | B1 | 1.1b |
\(\displaystyle 40000\left[0.032\frac{\mathrm{d}x}{\mathrm{d}t} - 0.025\frac{\mathrm{d}y}{\mathrm{d}t}\right] + 2800[0.032x - 0.025y] + \frac{13}{0.025}\left[0.032x - \frac{\mathrm{d}y}{\mathrm{d}t}\right]\) \(\displaystyle = 1280\frac{\mathrm{d}x}{\mathrm{d}t} - 1000\frac{\mathrm{d}y}{\mathrm{d}t} + 89.6x - 70y + 16.64x - 520\frac{\mathrm{d}y}{\mathrm{d}t}\) \(= 1280[0.025y - 0.045x + 2] - 1520[0.032x - 0.025y] + 89.6x - 70y + 16.64x\) \(= A\) (corrected from the printed mark scheme: the terms \(+ 89.6x - 70y + 16.64x\) are missing from the third line) | M1 | 1.1b |
| \(32y - 57.6x + 2560 - 48.64x + 38y - 70y + 106.24x = 2560\,*\) (corrected from the printed mark scheme: \(106.24x\) is printed as \(106.24\)) | A1* | 2.1 |
| (3) |
Notes
B1: Differentiates the second equation with respect to \(t\) correctly. May have rearranged to make \(x\) the subject first. The dot notation for derivatives may be used.
M1: Uses the second equation to eliminate \(x\) to achieve an equation in \(y\), \(\dfrac{\mathrm{d}y}{\mathrm{d}t}, \dfrac{\mathrm{d}^2y}{\mathrm{d}t^2}\).
A1*: Achieves the printed answer with no errors, allow dot notation
\[40\,000\ddot{y} + 2800\dot{y} + 13y = 2560\]Alternative
B1: Differentiates the second equation with respect to \(t\) correctly. May have rearranged to make \(x\) the subject first. The dot notation for derivatives may be used.
M1: Substitutes in the printed differential equation uses both equations to remove all derivative, form an expression involving \(x\)’s and \(y\)’s which simplifies to a constant.
A1*: Achieves 2560 with no errors seen
| Scheme | Marks | AO |
|---|---|---|
| \(40000m^2 + 2800m + 13\{= 0\} \Rightarrow m = \ldots\) | M1 | 3.4 |
| \(\text{CF}: y = A\mathrm{e}^{m_1t} + B\mathrm{e}^{m_2t}\) | M1 | 1.1b |
| \(\text{CF}: y = A\mathrm{e}^{-\frac{t}{200}} + B\mathrm{e}^{-\frac{13t}{200}}\) \(\text{CF}: y = A\mathrm{e}^{-0.005t} + B\mathrm{e}^{-0.065t}\) | A1 | 1.1b |
| PI: Try \(y = k \Rightarrow 13k = 2560 \Rightarrow k = \ldots\left\{\dfrac{2560}{13}\right\}\) | M1 | 3.4 |
| GS: \(y = A\mathrm{e}^{-\frac{t}{200}} + B\mathrm{e}^{-\frac{13t}{200}} + \dfrac{2560}{13}\) \(GS: y = A\mathrm{e}^{-0.005t} + B\mathrm{e}^{-0.065t} + \dfrac{2560}{13}\) | A1ft | 1.1b |
| (5) |
Notes
M1: Uses the model to form and attempt to solve the auxiliary equation the PI (Accept a correct equation followed by two values for \(m\) as an attempt to solve.)
M1: Forms the complementary function correct for their roots (so if repeated or complex roots found, award for appropriate form for CF). Must be in terms of \(t\) only (not \(x\))
A1: Correct CF
M1: Chooses the correct form of the PI according to the model and uses a complete method to find the PI
A1ft: Combines their CF (which need not be correct) with the correct PI to give \(y\) in terms of \(t\) so look for \(y\) = their CF + \(\dfrac{2560}{13}\), accepting awrt 197
| Scheme | Marks | AO |
|---|---|---|
| \(t = 0, y = 0 \Rightarrow 0 = A + B + \dfrac{2560}{13}\) | M1 | 3.4 |
| \(t = 0, y = 0, x = 0 \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}t} = 0.032 \times 0 - 0.025 \times 0 = 0\) Or Used \(x = \dfrac{1}{0.032}\left(\dfrac{\mathrm{d}y}{\mathrm{d}t} + 0.025y\right)\) to find an equation in \(t\) | B1 | 3.4 |
| \(\Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}t} = -\dfrac{A}{200}\mathrm{e}^{-\frac{t}{200}} - \dfrac{13B}{200}\mathrm{e}^{-\frac{13t}{200}} = 0 \Rightarrow -\dfrac{A}{200} - \dfrac{13B}{200} = 0 \Rightarrow A = -13B\) Or \(x = \dfrac{1}{0.032}\left[-0.005A\mathrm{e}^{-\frac{t}{200}} - 0.065B\mathrm{e}^{-\frac{13t}{200}} + 0.025\left(A\mathrm{e}^{-\frac{t}{200}} + B\mathrm{e}^{-\frac{13t}{200}} + \dfrac{2560}{13}\right)\right]\) \(x = \dfrac{5}{8}A\mathrm{e}^{-\frac{t}{200}} - \dfrac{5}{4}B\mathrm{e}^{-\frac{13t}{200}} + \dfrac{2000}{13} \Rightarrow 0 = \dfrac{5}{8}A - \dfrac{5}{4}B + \dfrac{2000}{13}\) | M1 | 1.1b |
| \(y = -\dfrac{640}{3}\mathrm{e}^{-\frac{t}{200}} + \dfrac{640}{39}\mathrm{e}^{-\frac{13t}{200}} + \dfrac{2560}{13}\) | A1 | 1.1b |
| (4) |
Notes
M1: Uses the initial conditions of the model to set up an equation in \(A\) and \(B\) from their general solution.
B1: Uses the initial conditions of the model to find the value of \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) when \(t = 0\). This can be implied.
Alternatively uses \(x = \dfrac{1}{0.032}\left(\dfrac{\mathrm{d}y}{\mathrm{d}t} + 0.025y\right)\) to find an equation in \(A\) and \(B\) from
M1: Differentiates their general solution, substitutes \(t = 0\) and sets equal to their initial value of \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) to form another equation in \(A\) and \(B\) and proceed at least as far as finding \(A\) in terms of \(B\) oe.
Alternative substitutes \(y\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) into their \(x\) equation and uses \(x = 0\) when \(t = 0\) to find an equation in \(A\) and \(B\)
A1: Correct particular solution, accepting awrt 197
| Scheme | Marks | AO |
|---|---|---|
| As \(t \rightarrow \infty, \mathrm{e}^{-kt} \rightarrow 0\) for \(k \gt 0\) so \(y \rightarrow \ldots\), | M1 | 1.1b |
| \(y \rightarrow \dfrac{2560}{13} \approx 196\) or \(197\) so the rate of administration is sufficient to reach the required level. | A1ft | 3.2b |
| (2) | ||
| (14 marks) |
Notes
M1: Uses the limit of the exponential terms is zero to find the long term limit of the concentration
A1ft: Follow through on their constant term and draws a relevant conclusion.